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Let \(D = \left\{ {{{\mathop{\rm d}\nolimits} _1},{{\mathop{\rm d}\nolimits} _2},{{\mathop{\rm d}\nolimits} _3}} \right\}\) and \(F = \left\{ {{{\mathop{\rm f}\nolimits} _1},{{\mathop{\rm f}\nolimits} _2},{{\mathop{\rm f}\nolimits} _3}} \right\}\) be bases for a vector space \(V\), and suppose \({{\mathop{\rm f}\nolimits} _1} = 2{{\mathop{\rm d}\nolimits} _1} - {{\mathop{\rm d}\nolimits} _2} + {{\mathop{\rm d}\nolimits} _3}\), \({{\mathop{\rm f}\nolimits} _2} = 3{{\mathop{\rm d}\nolimits} _2} + {{\mathop{\rm d}\nolimits} _3}\) and \({{\mathop{\rm f}\nolimits} _3} = - 3{{\mathop{\rm d}\nolimits} _1} + 2{{\mathop{\rm d}\nolimits} _3}\).

a. Find the change-of-coordinates matrix from \(F\) to \(D\).

b. Find \({\left[ {\mathop{\rm x}\nolimits} \right]_D}\) for \({\mathop{\rm x}\nolimits} = {{\mathop{\rm f}\nolimits} _1} - 2{{\mathop{\rm f}\nolimits} _2} + 2{{\mathop{\rm f}\nolimits} _3}\).

Short Answer

Expert verified
  1. The change-of-coordinates matrix from \(F\) to \(D\) is \(\mathop P\limits_{D \leftarrow F} = \left[ {\begin{array}{*{20}{c}}2&0&{ - 3}\\{ - 1}&3&0\\1&1&2\end{array}} \right]\).
  2. The \(D - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_D} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 7}\\3\end{array}} \right]\).

Step by step solution

01

State the change-of-coordinate matrix

Let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},...,{{\mathop{\rm c}\nolimits} _n}} \right\}\) be bases of a vector space \(V\). Then according to Theorem 15,there is a unique \(n \times n\) matrix \(\mathop P\limits_{C \leftarrow B} \) such that \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \mathop P\limits_{C \leftarrow B} {\left[ {\mathop{\rm x}\nolimits} \right]_B}\).

The columns of \(\mathop P\limits_{C \leftarrow B} \) are the \(C - \)coordinate vectors of the vectors in the basis \(B\). That is, \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]}_C}}&{{{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]}_C}}& \cdots &{{{\left[ {{{\mathop{\rm b}\nolimits} _n}} \right]}_C}}\end{array}} \right]\).

02

Determine the change-of-coordinate matrix from \(F\) to \(D\)

a)

It is given that \({{\mathop{\rm f}\nolimits} _1} = 2{{\mathop{\rm d}\nolimits} _1} - {{\mathop{\rm d}\nolimits} _2} + {{\mathop{\rm d}\nolimits} _3},{{\mathop{\rm f}\nolimits} _2} = 3{{\mathop{\rm d}\nolimits} _2} + {{\mathop{\rm d}\nolimits} _3}\ and \({{\mathop{\rm f}\nolimits} _3} = - 3{{\mathop{\rm d}\nolimits} _1} + 2{{\mathop{\rm d}\nolimits} _3}\). Then, \({\left[ {{{\mathop{\rm f}\nolimits} _1}} \right]_D} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\1\end{array}} \right],{\left[ {{{\mathop{\rm f}\nolimits} _2}} \right]_D} = \left[ {\begin{array}{*{20}{c}}0\\3\\1\end{array}} \right],{\left[ {{{\mathop{\rm f}\nolimits} _3}} \right]_D} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\0\\2\end{array}} \right]\).

\(\begin{aligned} \mathop P\limits_{D \leftarrow F} &= \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm f}\nolimits} _1}} \right]}_D}}&{{{\left[ {{{\mathop{\rm f}\nolimits} _2}} \right]}_D}}&{{{\left[ {{{\mathop{\rm f}\nolimits} _3}} \right]}_D}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}2&0&{ - 3}\\{ - 1}&3&0\\1&1&2\end{array}} \right]\end{aligned}\)

Thus, the change-of-coordinates matrix from \(F\) to \(D\) is \(\mathop P\limits_{D \leftarrow F} = \left[ {\begin{array}{*{20}{c}}2&0&{ - 3}\\{ - 1}&3&0\\1&1&2\end{array}} \right]\).

03

Determine \({\left[ {\mathop{\rm x}\nolimits}   \right]_D}\) for \({\mathop{\rm x}\nolimits}  = {{\mathop{\rm f}\nolimits} _1} - 2{{\mathop{\rm f}\nolimits} _2} + 2{{\mathop{\rm f}\nolimits} _3}\)

b)

It is given that \({\mathop{\rm x}\nolimits} = {{\mathop{\rm f}\nolimits} _1} - 2{{\mathop{\rm f}\nolimits} _2} + 2{{\mathop{\rm f}\nolimits} _3}\), then \({\left[ {\mathop{\rm x}\nolimits} \right]_F} = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\2\end{array}} \right]\).

Use part (a) to compute the D-coordinate vector.

\(\begin{aligned} {\left[ {\mathop{\rm x}\nolimits} \right]_D} &= \mathop P\limits_{D \leftarrow F} {\left[ {\mathop{\rm x}\nolimits} \right]_F}\\ &= \left[ {\begin{array}{*{20}{c}}2&0&{ - 3}\\{ - 1}&3&0\\1&1&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{2 + 0 - 6}\\{ - 1 - 6 + 0}\\{1 - 2 + 4}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 7}\\3\end{array}} \right]\end{aligned}\)

Therefore, the \(D - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_D} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 7}\\3\end{array}} \right]\).

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Most popular questions from this chapter

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss.

Question 18: Suppose A is a \(4 \times 4\) matrix and B is a \(4 \times 2\) matrix, and let \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) represent a sequence of input vectors in \({\mathbb{R}^2}\).

  1. Set \({{\mathop{\rm x}\nolimits} _0} = 0\), compute \({{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}\) from equation (1), and write a formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix \(M\) appearing in equation (2). (Note: The matrix \(M\) is constructed as a partitioned matrix. Its overall size here is \(4 \times 8\).)
  2. Suppose \(\left( {A,B} \right)\) is controllable and v is any vector in \({\mathbb{R}^4}\). Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) such that \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

20. \(A = \left( {\begin{array}{*{20}{c}}{.8}&{ - .3}&0\\{.2}&{.5}&1\\0&0&{ - .5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right)\).

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